Let `f(x)=a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x ,` where `a_i ' s` are real and `f(x)=0` has a positive root `alpha_0dot` Then `f^(prime)(x)=0` has a positive root `alpha_1` such that `0

Options

A. f(x)=0 has a root `alpha_(1)` such that `0lt alpha_(1)ltalpha_(0)`
B. f(x) =0 has at least two real roots
C. f(x) =00 has at least one real root
D. none of these

Answer

1,2,3

Explanation

`f(x)=a_(5)x^(5)+a_(4)x^(4)+a_(3)x^(3)+a_(2)x^(2)+a_(1)x`
Thus f(x) =0 has one root x=0
Aslo given that f(X) =0 has positive root `a_(0)` thus the equation must have at leat three lreal roots (as complex root occurs in conjugate pair) Thus f(X) =0 has at least two real roots as between two roots of f(X) =0 there lies at leat one root of f(x)=0
Similarly we can say that f(x) =0 has at least one real root further f(x)=0 has one root between roots x=0 and `x=a_(0)` of f(X)=0